Let f: N (0, ) be a function satisfying f(x+y) = f(x)f(y) for all x, y N . If _ k=1 ⁹ f(k) = 1024 _ k=1 ⁹ 1 f(k) , then the value of f(4) is

Step-by-step solution

Given f(x+y) = f(x)f(y) for all x, y N . By substituting x=1, y=1 , we get f(2) = (f(1))^2 . By induction, f(k) = a^k for all k N , where a = f(1) > 0 . The sum of the first 9 terms is: _ k=1 ⁹ f(k) = a + a^2 + + a^9 = a ( a^9 - 1 a - 1 ) The sum of the reciprocals is: _ k=1 ⁹ 1 f(k) = 1 a + 1 a^2 + + 1 a^9 This is a geometric progression with first term 1 a and common ratio 1 a : _ k=1 ⁹ 1 f(k) = 1 a ( 1 - ( 1 a )^9 1 - 1 a ) = 1 a ( a^9 - 1 a^9 a - 1 a ) = a^9 - 1 a^9 (a - 1) According to the given condition: a (